Transcript Section 5.7 Testing Hypotheses about Population Means

Lecture Unit 5
Section 5.7
Testing Hypotheses
about Means
1
Sweetness in cola soft drinks
Cola manufacturers want to test how much the sweetness of cola drinks is
affected by storage. The sweetness loss due to storage was evaluated by 10
professional tasters by comparing the sweetness before and after storage
(a positive value indicates a loss of sweetness):










Taster
Sweetness loss
1
2
3
4
5
6
7
8
9
10
2.0
0.4
0.7
2.0
−0.4
2.2
−1.3
1.2
1.1
2.3
We want to test if storage results
in a loss of sweetness, thus:
H0: m = 0 versus HA: m > 0
where m is the mean sweetness
loss due to storage.
We also do not know the population parameter s, the standard deviation of the
sweetness loss.
As in hypothesis tests for p, a hypothesis test for m
requires a few steps:
1. State the null and alternative hypotheses (H0 versus HA)
a) Decide on a one-sided or two-sided test
2. Calculate the test statistic t and determining its degrees of freedom
3. Find the area under the t distribution with the t-table or
technology
4. State the P-value (or find bounds on the P-value) and interpret the
result
As in hypothesis tests for a population proportion p, a
hypothesis test for a population mean m requires a few
steps:
1. State the null and alternative hypotheses (H0 versus HA)
a) Decide on a one-sided or two-sided test
H0: m = m0 versus HA: m > m0 (1 –tail test)
H0: m = m0 versus HA: m < m0 (1 –tail test)
H0: m = m0 versus HA: m ≠ m0 (2 –tail test)
We perform a hypothesis test with null hypothesis
H : m = m0
using the test statistic
y  m0
t
SE ( y )
where the standard error of
y
is .
s
SE ( y ) 
n
When the null hypothesis is true, the test statistic follows a
t distribution with n-1 degrees of freedom. We use that
model to obtain a P-value.
The one-sample t-test; P-Values
Recall:
The P-value is the probability, calculated assuming the null
hypothesis H0 is true, of observing a value of the test statistic
more extreme than the value we actually observed.
The calculation of the P-value depends on whether the
hypothesis test is 1-tailed
(that is, the alternative hypothesis is
HA :m < m0 or HA : m > m0)
or 2-tailed
(that is, the alternative hypothesis is HA : m ≠ m0).
6
P-Values
Assume the value of the test statistic t is t0
If HA: m > m0, then P-value=P(t > t0)
If HA: m < m0, then P-value=P(t < t0)
If HA: m ≠ m0, then P-value=2P(t > |t0|)
7
Sweetening colas (continued)
Is there evidence that storage results in sweetness loss in colas?
H0: m = 0 versus Ha: m > 0 (one-sided test)
t
y  m0
s
1.02  0

 2.70
n 1.196 10
P  value  P(t9  2.70)
Conf. Level
Two Tail
One Tail
df
9
0.1
0.9
0.45
0.3
0.7
0.35
0.5
0.5
0.25
0.1293
0.3979
0.7027
0.7
0.3
0.15
0.8
0.9
0.2
0.1
0.1
0.05
Values of t
1.0997 1.3830 1.8331
0.95
0.05
0.025
0.98
0.02
0.01
0.99
0.01
0.005
2.2622
2.8214
3.2498
Taster
Sweetness loss
1
2.0
2
0.4
3
0.7
4
2.0
5
-0.4
6
2.2
7
-1.3
8
1.2
9
1.1
10
2.3
___________________________
Average
1.02
Standard deviation
1.196
Degrees of freedom
n−1=9
2.2622 < t = 2.70 < 2.8214; thus 0.01 < P-value < 0.025.
Since P-value < .05, we reject H0. There is a significant loss
of sweetness, on average, following storage.
TDIST(x, degrees_freedom, tails)
TDIST = P(t > x) for a random variable t following the t distribution (x positive).
Use it in place of t-table to obtain the P-value.
◦ x is the absolute value of the test statistic.
◦ Deg_freedom is an integer indicating the number of degrees of freedom.
◦ Tails specifies the number of distribution tails to return. If tails = 1, TDIST returns
the one-tailed P-value. If tails = 2, TDIST returns the two-tailed P-value.
t
y  m0
s
1.02  0

 2.70
n 1.196 10
2.2622 < t = 2.70 < 2.8214; thus 0.01 < p < 0.025.
10
The NYC Visitors Bureau
claims that the average
cost of a hotel room is
$168 per night. A
random sample of 25
hotels resulted in
y = $172.50 and
s = $15.40.
H0: μ = 168
HA: μ  168
t, 24 df
H0: μ = 168
HA: μ  168
.079
.079
 n = 25; df = 24
y  $172.50, s  $15.40
t 
-1. 46
yμ
172.50  168

 1.46
s
15.40
n
25
Conf. Level
Two Tail
One Tail
df
24
0.1
0.9
0.45
0.3
0.7
0.35
0.5
0.5
0.25
0.1270
0.3900
0.6848
0
0.7
0.3
0.15
0.8
0.9
0.2
0.1
0.1
0.05
Values of t
1.0593 1.3178 1.7109
1. 46
P-value = .158
P  value  2P(t  1.46)
0.95
0.05
0.025
0.98
0.02
0.01
0.99
0.01
0.005
2.0639
2.4922
2.7969
Do not reject H0: not sufficient evidence that
true mean cost is different than $168
A popcorn maker wants a combination of
microwave time and power that delivers highquality popped corn with less than 10%
unpopped kernels, on average. After testing, the
research department determines that power 9 at 4
minutes is optimum. The company president
tests 8 bags in his office microwave and finds the
following percentages of unpopped kernels: 7,
13.2, 10, 6, 7.8, 2.8, 2.2, 5.2.
Do the data provide evidence that the mean
percentage of unpopped kernels is less than 10%?
H0: μ = 10
HA: μ < 10
where μ is true unknown mean percentage of unpopped
kernels
t, 7 df
H0: μ = 10
HA: μ < 10
.02
 n = 8; df = 7
y  6.775, s  3.64
t 
-2. 51
ym
6.775  10

 2.51
s
3.64
n
8
Conf. Level
Two Tail
One Tail
df
7
0
0.1
0.9
0.45
0.3
0.7
0.35
0.5
0.5
0.25
0.1303
0.4015
0.7111
Exact P-value = .02
P  value  P(t < 2.51)
0.7
0.3
0.15
0.8
0.9
0.2
0.1
0.1
0.05
Values of t
1.1192 1.4149 1.8946
0.95
0.05
0.025
0.98
0.02
0.01
0.99
0.01
0.005
2.3646
2.9980
3.4995
Reject H0: there is sufficient evidence that true
mean percentage of unpopped kernels is less than
10%